Minimal Polynomial

Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation.

$$\mathbf{A}^n + d_1\mathbf{A}^{n-1} + \cdots + d_{n-1}\mathbf{A} + d_n\mathbf{I} = 0$$

Remark: Cayley-Hamilton Theorem basically states that nth and higher powers of a matrix can be expressed in terms of lower powers of the matrix. This is a very useful property in solving linear systems.

$A^n = -d_1A^{n-1} - \cdots - d_{n-1}A - d_nI$

For previous example, we have:

$A^2 - 4A + 3I = 0$

$A^2 = 4A - 3I$

$A^3 = 4A^2 - 3A = 4(4A - 3I) - 3A = 13A - 12I$

$A^4 = 4A^3 - 3A^2 = 4(13A - 12I) - 3(4A - 3I) = 49A - 48I$

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$\large exp(s) = \sum_{n=0}^{\infty} \frac{s^n}{n!} = I + s + \frac{s^2}{2!} + \frac{s^3}{3!} + \cdots$

$\large exp(A) = \sum_{n=0}^{\infty} \frac{A^n}{n!} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots$

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$A^{-1} = ?$

$A^2 - 4A + 3I = 0$

$A - 4I + 3A^{-1} = 0$

$A^{-1} = \Large \frac{(4I - A)}{3}$

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Example: $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ $A^{100}=?$

Solution:

$A^{100} = \alpha A + \beta I$

$\lambda_1 = 3, \lambda_2 = 1 \text{, and } e_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, e_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

$A e_1 = \lambda_1 e_1 \Rightarrow A^2 e_1 = \lambda_1^2 e_1 \Rightarrow A^3 e_1 = \lambda_1^3 e_1 \Rightarrow \cdots \Rightarrow A^{100} e_1 = \lambda_1^{100} e_1$

$A e_2 = \lambda_2 e_2 \Rightarrow A^2 e_2 = \lambda_2^2 e_2 \Rightarrow A^3 e_2 = \lambda_2^3 e_2 \Rightarrow \cdots \Rightarrow A^{100} e_2 = \lambda_2^{100} e_2$

$A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_1 = \alpha A e_1 + \beta e_1 \Rightarrow \lambda_1^{100} e_1 = \alpha \lambda_1 e_1 + \beta e_1 \Rightarrow \lambda_1^{100} = \alpha \lambda_1 + \beta \implies 3^{100} = 3\alpha + \beta$

$A^{100} = \alpha A + \beta I \Rightarrow A^{100} e_2 = \alpha A e_2 + \beta e_2 \Rightarrow \lambda_2^{100} e_2 = \alpha \lambda_2 e_2 + \beta e_2 \Rightarrow \lambda_2^{100} = \alpha \lambda_2 + \beta \implies 1^{100} = \alpha + \beta$

$\begin{cases} 3^{100} = 3\alpha + \beta \\ 1 = \alpha + \beta \end{cases} \Rightarrow \begin{matrix} \alpha = \frac{3^{100} - 1}{2} \\ \beta = \frac{-3^{100} + 3}{2} \end{matrix}$

$A^{100} = \frac{3^{100} - 1}{2} A + \frac{-3^{100} + 3}{2} I$

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Minimal Polynomial

Definition: The monic polynomial is the polynomial with the highest degree coefficient equal to 1.

$s^n + a_1s^{n-1} + \cdots \rightarrow$ monic

$2s^n + a_1s^{n-1} + \cdots \rightarrow$ not monic

Definition: The minimal polynomial of a matrix is the monic polynomial of the lowest degree that has the matrix as a root.

$$m(A) = 0_{n \times n}$$

Theorem: Given a matrix $A = \mathbb{C}^{n \ times n}$, let m(s) be its minimal polynomial. Then, the minimal polynomial is the smallest degree polynomial that satisfies the following:

i) m(s) is unique. ii) m(s) divides d(s) with no remainder.

$$\exists q(s) \text{ s.t. } d(s) = q(s)m(s)$$

iii) Every root of m(s) is a root of d(s).

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Example: $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ $\Large \substack{d(s) = (s-1)(s-2)(s-3) \\ m(s) = (s-1)(s-2)(s-3)}$

Remark: When A has distinct eigenvalues (which implies diagonalizability but converse is not true), the minimal polynomial is the same as the characteristic polynomial.

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Example: $A_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ $\Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)} \ \ \ \ \ \ \ \normalsize A_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$ $\Large \substack{d(s) = (s-1)(s-2)^2 \\ m(s) = (s-1)(s-2)^2}$

A way to check if a minimal polynomial is correct is to check if the characteristic polynomial is zero when the minimal polynomial is substituted for s.

Let $m_1(s) = (s-1)(s-2)$ then $m_1(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark$

Let $m_2(s) = (s-1)(s-2)$ then $m_2(A) = (A-1I)(A-2I) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \neq 0$

Then, $m_2(s) = (s-1)(s-2)^2 \implies m_2(A) = (A-1I)(A-2I)^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \checkmark$

$\mathbb{R}^3 = N(A-1I) \oplus N((A-2I)^2)$ where,

$N(A-1I) = N \bigg ( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N(A-1I) \bigg ) = 1$

$N((A-2I)^2) = N \bigg ( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \bigg ) = \text{span} \bigg \{ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg \} \ \ \text{dim} \bigg ( N((A-2I)^2) \bigg ) = 2$

Remark: There is a relationship between decomposition form and minimal polynomial in terms of power.

$$m(s) = (s-\lambda_1)^{d_1} \cdots (s-\lambda_k)^{d_k}$$

$$\mathbb{R}^n = N(A-\lambda_1I)^{d_1} \oplus \cdots \oplus N(A-\lambda_kI)^{d_k}$$

Theorem:

$$\mathbb{C}^n = N(A-\lambda_1I)^{m_1} \oplus N(A-\lambda_2I)^{m_2} \oplus \cdots \oplus N(A-\lambda_kI)^{m_k}$$

$$d(s) = (s-\lambda_1)^{r_1} \cdots (s-\lambda_k)^{r_k}$$

$$r_1 + r_2 + \cdots + r_k = n$$

$$m(s) = (s-\lambda_1)^{m_1} \cdots (s-\lambda_k)^{m_k}$$

$$1 \leq m_i \leq r_i$$

$\bar A = \begin{bmatrix} \bar A_1 & 0 & \cdots & 0 \\ 0 & \bar A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \bar A_k \end{bmatrix}$ $\Large \substack{ \bar A = B^{-1}AB \text{, B is composed of the basis vectors} \\ B = \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix} \text{ for the } N(A-\lambda_iI)^{m_i} }$

Size of $\bar A_i$ is $\text{dim} \bigg ( N(A-\lambda_iI)^{m_i} \bigg )$

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Example: $A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}$ $\Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^2(s-2)}$

Solution: Let $\Sigma_1$ be $A-\lambda_1I$ and $\Sigma_2$ be $A-\lambda_2I$.

$\Sigma_1 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 2 \neq 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)}= 4 - 2 = 2$

Then we need to check the dimension of $N(\Sigma_1^2)$.

$\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 1 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^2)} = 4 - 1 = 3 = r_1$

Lets briefly check the dimension of $N(\Sigma_1^3)$.

$\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 4 - 1 = 3 = r_1$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$

$\Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 4 - 1 = 3 = r_1$

Lets check the second eigenvalue.

$\Sigma_2 = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1$

Then the minimal polynomial is $m(s) = (s-1)^2(s-2)$.

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Example: $A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}$ $\Large \substack{d(s) = (s-1)^3(s-2) \\ m(s) = (s-1)^3(s-2)}$

Solution: Let $\Sigma_1$ be $A-\lambda_1I$ and $\Sigma_2$ be $A-\lambda_2I$.

$\Sigma_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 4 - 3 = 1$

Then we need to check the dimension of $N(\Sigma_1^2)$.

$\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 4 - 2 = 2$

Check the dimension of $N(\Sigma_1^3)$.

$\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)}= 4 - 1 = 3$

Now we can check the second eigenvalue.

$\Sigma_2 = \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 4 - 3 = 1$

Then the minimal polynomial is $m(s) = (s-1)^3(s-2)$.

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Example: $A = \begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}$ $\Large \substack{d(s) = (s-2)^7(s-3)^3 \\ m(s) = ?}$

Solution: For each jordan block, we need to check the dimension of $N(\Sigma_1^i)$.

The largest jordan block will have the largest dimension of $N(\Sigma_1^i)$. The rest of the jordan blocks will have dimension of $N(\Sigma_1^i)$ equal to the size of the jordan block. Hence the geometric multiplicity wont increase.

Let $\Sigma_1$ be $A-\lambda_1I$ and $\Sigma_2$ be $A-\lambda_2I$. Check the largest jordan block.

$\Sigma_1 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1) \bigg ) = 1 \neq r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1)} = 3 - 2 = 1$

Then we need to check the dimension of $N(\Sigma_1^2)$.

$\Sigma_1^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^2) \bigg ) = 2 \neq r_1 \rightarrow \text{dim}{V}- \text{dim}{R(\Sigma_1^2)} = 3 - 1 = 2$

Check the dimension of $N(\Sigma_1^3)$.

$\Sigma_1^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^3) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^3)} = 3 - 0 = 3$

We need to continue to see that further powers of $\Sigma_1$ will not increase the dimension of $N(\Sigma_1^i)$.

$\Sigma_1^4 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_1^4) \bigg ) = 3 = r_1 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_1^4)} = 3 - 0 = 3$

This implies that the further jordan blocks will not increase the dimension of $N(\Sigma_1^i)$. Hence they are not needed to be checked for minimal polynomial.

Now we can check the second eigenvalue.

$\Sigma_2 = \begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2) \bigg ) = 1 \neq r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2)} = 2 - 1 = 1$

$\Sigma_2^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{dim} \bigg ( N(\Sigma_2^2) \bigg ) = 2 = r_2 \rightarrow \text{dim}{V} - \text{dim}{R(\Sigma_2^2)} = 2 - 0 = 2$

Then the minimal polynomial is $m(s) = (s-2)^7(s-3)^2$.

Remark:

$$\text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 1}$$

$$\text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 2}$$

$$\text{dim} \bigg ( N(A-\lambda_iI)^3 \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ 3}$$

$$\text{dim} \bigg ( N(A-\lambda_iI)^k \bigg ) - \text{dim} \bigg ( N(A-\lambda_iI)^{k-1} \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_i \text{ with size $\geq$ k}$$

Example: $A \in \mathbb{R}^4$ Which has a single eigenvalue $\lambda_1 = 7$ and the geometric multiplicity is 2. Find all possible jordan forms.

Solution: The characteristic polynomial is $d(s) = (s-7)^4$ and the minimal polynomial is $m(s) = (s-7)^2$. This suggests that,

$\text{dim} \bigg ( N(A-7I)^2 \bigg ) = 4 = r_1 \rightarrow \text{dim}(V) = \text{dim} \bigg ( N(A-7I)^2 \bigg )$

Then for the $\text{dim}\bigg (N(A-7I)\bigg )$ we have 3 possibilities, 1, 2, or 3.

$\text{a) Let } \text{dim} N(A-7I) = 3$

$\ \ \ \ \ \text{\# of jordan blocks } = 3$

$\ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 3 = 1$

$\ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = 3 - 3 = 0$

$A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 7 \end{bmatrix}$

$\text{a) Let } \text{dim} N(A-7I) = 2$

$\ \ \ \ \ \text{\# of jordan blocks } = 2$

$\ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 2 = 2$

$\ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 3 = \text{dim}\bigg ( N(A-7I)^3 \bigg ) - \text{ dim}\bigg ( N(A-7I)^2 \bigg ) = 4 - 4 = 0$

$A = \begin{bmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 1 \\ 0 & 0 & 0 & 7 \end{bmatrix}$

$\text{a) Let } \text{dim} N(A-7I) = 1$

$\ \ \ \ \ \text{\# of jordan blocks } = 1$

$\ \ \ \ \ \text{\# of jordan blocks w/ size } \geq 2 = 4 - 1 = 3$

However, this is not possible since there cannot be 3 jordan blocks with size greater than or equal to 2.

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Example: Suppose a matrix $A \in \mathbb{R}^{8\times 8}$ has the following subspace dimensions:

$$\text{dim} \bigg ( N(A-3I) \bigg ) = 5 \\ \text{dim} \bigg ( N(A-3I)^2 \bigg ) = 7 \\ \text{dim} \bigg ( N(A-3I)^3 \bigg ) = 8$$

a) Find the characteristic polynomial of A.

b) Find the minimal polynomial of A.

c) Find the possible Jordan forms of A.

Solution:

a) We know that the characteristic polynomial is $d(s) = (s-3)^8$.

b) We know that the minimal polynomial is $m(s) = (s-3)^3$ since the dimension of $N(A-3I)^3$ is same as the dimension of the vector space. This implies that the geometric multiplicity is 3.

c) Starting from the largest jordan block, having geometric multiplicity 3 suggest that the largest jordan block has size 3.

$\text{dim} \bigg ( N(A-3I)^3 \bigg ) - \text{dim} \bigg ( N(A-3I)^2 \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 3 = 8 - 7 = 1$

$\text{dim} \bigg ( N(A-3I)^2 \bigg ) - \text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 2 = 7 - 5 = 2$

We have 2 that are greater than or equal to 2. This implies that we have 1 jordan block with size 3 and 1 jordan blocks with size 2.

$\text{dim} \bigg ( N(A-3I) \bigg ) = \text{\# of jordan blocks corresponding to } \lambda_1 \text{ with size }\geq 1 = 5$

This implies that we have 3 jordan blocks with size 1.

$A = \begin{bmatrix}3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}$

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#EE501 - Linear Systems Theory at METU